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Math Help - Maclaurin Series(approximation)

  1. #1
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    Maclaurin Series(approximation)

    The Maclaurin series for sin x is

    \frac {\sin x}x = \sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k x^{2k} }}<br />
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..



    use the corresponding Maclaurin polynomial of degree 5 to approximate

    \int\frac{\sin x}{x} dx

    integration from -1 to 1

    i dont have any idea to solve it..
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  2. #2
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    Quote Originally Posted by nameck View Post
    The Maclaurin series for sin x is

    \frac {\sin x}x = \sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k x^{2k} }}<br />
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..



    use the corresponding Maclaurin polynomial of degree 5 to approximate

    \int\frac{\sin x}{x} dx

    integration from -1 to 1

    i dont have any idea to solve it..
    Integrate "term by term"!

    Since \frac {\sin x}x = \sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k x^{2k} }}<br />
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..

    \int_{-1}^1 \frac{sin x}{x}= \int_{-1}^1 dx- \frac{1}{3!}\int_{-1}^1 x^2 dx+ \frac{1}{5!}\int_{-1}^1 x^4 dx

    Since the problem said "polynomial of degree 5" and integrating x^4 will give degree 5, you can stop there.
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  3. #3
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    got it!! =)
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  4. #4
    MHF Contributor chisigma's Avatar
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    The McLaurin expasion of the sin function is...

    \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}\cdot x^{2n+1} (1)

    Only the 'odd' powers of x give contribution, so that is...

    \int_{-a}^{a} \sin x\cdot dx =0 (2)

    ... no matter which is a...



    Merry Christmas from Italy

    \chi \sigma
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  5. #5
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    Yes, but this was sin(x)/x which is an even function.

    Quote Originally Posted by chisigma View Post
    The McLaurin expasion of the sin function is...

    \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}\cdot x^{2n+1} (1)

    Only the 'odd' powers of x give contribution, so that is...

    \int_{-a}^{a} \sin x\cdot dx =0 (2)

    ... no matter which is a...



    Merry Christmas from Italy

    \chi \sigma
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  6. #6
    MHF Contributor chisigma's Avatar
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    I have posted that in order to clarify the misundestanding caused by the fact that the function to be integrated is \frac{\sin x}{x} and not \sin x as written is the first post by nameck: the Maclaurin series for sin x is...



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  7. #7
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    i got 1.8922222..
    is my answer correct?
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